如何使用Kotlin就地过滤列表? [英] How to filter a list in-place with Kotlin?

问题描述

在Java中，我可以使用以下代码从列表中删除项目:

In Java I can remove items from a list with this code:

private void filterList(List<Item> items) {
    Iterator<Item> iterator = items.iterator();
    while (iterator.hasNext()) {
        if (checkItem(iterator.next())) {
            iterator.remove();
        }
    }
}

如何在Kotlin中做到这一点(即在不重新创建的情况下删除List中的某些项目)?

How to make the same in Kotlin (i.e. remove some items in a List without re-creation)?

推荐答案

只需使用 .retainAll { ... } 或 .removeAll { ... } 都接受一个谓词，以对其进行就地过滤:

Just use .retainAll { ... } or .removeAll { ... }, both accepting a predicate, to filter it in-place:

items.retainAll { shouldRetain(it) }

items.removeAll { shouldRemove(it) }

请注意，为此，items应该是一个MutableList<T>，而不仅仅是List<T>，它是Kotlin中的只读列表，因此不会公开任何变异函数(请参阅:

Note that items should be a MutableList<T> for that, not just List<T>, which is a read-only list in Kotlin and thus does not expose any mutating functions (see: Collections in the language reference).

_{顺便说一句，对于支持随机访问的列表，有效地实现了这两个功能:然后，在删除每个项目之后，列表不会被压缩( O(n ² )) 时间最坏的情况下)，而是在处理列表时将其移至列表中，从而给 O(n)时间.}

_{By the way, these two function are implemented efficiently for lists that support random access: then the list is not compacted after each item is removed (O(n²) time worst-case), and instead the items are moved within the list as it is processed, giving O(n) time.}

如果您不想修改原始列表，则可以使用

And if you don't want to modify the original list, you can produce a separate collection with only the items you want to retain using .filter { ... } or .filterNot { ... }, this will work for read-only List<T> as well:

val filtered = items.filter { shouldRetain(it) }

val filtered = items.filterNot { shouldRemove(it) }

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